Find the maximum area and perimeter of inscribed rectangle of ellipse x 2 / a 2 + y 2 / B 2 = 1 (a > b > 0)

Find the maximum area and perimeter of inscribed rectangle of ellipse x 2 / a 2 + y 2 / B 2 = 1 (a > b > 0)

Let x = acost, y = bsint, so s = 4xy = 4absintcost = 2absin2t, when t is π / 4 + 2K π, the maximum area is 2Ab. C = 4 (acost + bsint) = 4 √ a ∧ 2 + B

There are four ellipses in a square. The side length of the square is 4cm. Find the area sum of the four ellipses

It's not the diagonal of an ellipse, but the figure of the intersection of four semicircles,
After adding four semicircles, subtracting the square area is the area of the intersecting figure
4*3.14*2^2/2-4^2=9.12cm^2

The side length of a square is 4cm. There is an ellipse in it. Find the area of the ellipse

S=π×a×b
So, s (area max of ellipse) = π R ^ 2 = 4 π
The solution set of the area of ellipse is (0,4 π]